Compute $Q_1$, $D_2$ and $P_{90}$ from the following data. | Marks | 10 | 20 | 30 | 40 | 50 | 60 | | --- | --- | --- | --- | --- | --- | --- | | No. of Students | 4 | 7 | 15 | 8 | 7 | 2 |

> > [!TLDR] > Building the cumulative frequency ($N=43$) and using $Q_1$: $(N+1)/4$-th, $D_2$: $2(N+1)/10$-th, $P_{90}$: $90(N+1)/100$-th items gives $Q_1 = 20$, $D_2 = 20$, $P_{90} = 50$. > > **Step 1 — Cumulative frequency table.** > > | Marks | Frequency | Cumulative freq. | > | --- | --- | --- | > | 10 | 4 | 4 | > | 20 | 7 | 11 | > | 30 | 15 | 26 | > | 40 | 8 | 34 | > | 50 | 7 | 41 | > | 60 | 2 | 43 | > > Total $N = 43$, so $N+1 = 44$. > > **Step 2 — First Quartile $Q_1$.** Size of $\dfrac{N+1}{4} = \dfrac{44}{4} = 11$-th item. The cumulative frequency reaches 11 at Marks $= 20$. So $Q_1 = 20$. > …

Compute Q₁, D₂ and P₉₀ from the following data. | – Class 11 Business Mathematics Statistics

Question 25 of 34

Compute $Q_1$ , $D_2$ and $P_{90}$ from the following data.

Marks	10	20	30	40	50	60
No. of Students	4	7	15	8	7	2

Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2023Subjective· 3mImportance★★★★★

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Building the cumulative frequency ( $N=43$ ) and using $Q_1$ : $(N+1)/4$ -th, $D_2$ : $2(N+1)/10$ -th, $P_{90}$ : $90(N+1)/100$ -th items gives $Q_1 = 20$ , $D_2 = 20$ , $P_{90} = 50$ .

Step 1 — Cumulative frequency table.

Marks	Frequency	Cumulative freq.
10	4	4
20	7	11
30	15	26
40	8	34
50	7	41
60	2	43

Total $N = 43$ , so $N+1 = 44$ .

Step 2 — First Quartile $Q_1$ . Size of $\dfrac{N+1}{4} = \dfrac{44}{4} = 11$ -th item. The cumulative frequency reaches 11 at Marks $= 20$ . So $Q_1 = 20$ .

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CA Foundation 2026Set jan-20261 markMCQ
$Q. If arithmetic mean of two numbers is 64 and harmonic mean is 16 then geometric mean is (A) 64 (B) 16 (C) 32 (D) 8$
›Reveal solutionSolution

Using the relation $GM^2 = AM\times HM$ , $GM=\sqrt{64\times16}=\sqrt{1024}=32$ .

Step 1 — recall the relation

For two positive numbers, $AM \times HM = GM^2$ .

Step 2 — substitute
$GM^2 = 64 \times 16 = 1024$
Step 3 — solve

$GM = \sqrt{1024} = 32$ …
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CA Foundation 2025Set jan-20251 markMCQ
$13, 8, 6, 3, 8, 13, 2x + 3, 8, 13, 3, 5, 7$
›Reveal solutionSolution

To make 13 the mode, set $2x+3=13\Rightarrow x=5$ , which raises 13's count to four (beating 8's three).

Step 1 — Count the fixed values

Data: $13, 8, 6, 3, 8, 13, 2x+3, 8, 13, 3, 5, 7$ .

Ignoring the unknown term: $13$ appears $3$ times, $8$ appears $3$ times, $3$ appears $2$ times; all others once.

Step 2 — Use the mode condition

The mode is given as $13$ . With $8$ also appearing three times, $13$ must occur MORE than $8$ , so the unknown term must itself be $13$ :
$2x+3 = 13 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5.$
Now $13$ appears four times — the unique mode. …
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CA Foundation 2025Set jan-20251 markMCQ
$Q. The best measure of central tendency is (A) Mean (B) Median (C) Mode (D) Range$
›Reveal solutionSolution

The arithmetic mean is the best measure of central tendency on the standard textbook criteria.

Step 1 — Criteria for an ideal average

An ideal measure of central tendency should be (i) rigidly defined, (ii) based on all observations, (iii) easy to understand and compute, (iv) capable of further algebraic treatment and (v) least affected by sampling fluctuations.

Step 2 — Why the mean wins

The arithmetic mean satisfies all these — every value enters the calculation and it feeds directly into higher statistics (combined mean, standard deviation, regression). It is therefore taken as the best/most widely used measure. …
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CA Foundation 2025Set may-20251 markMCQ
$Q. Which one of the following measures of central tendency is based on only fifty percent (50%) of the central values ? (A) Geometric Mean (B) Harmonic Mean (C) Median (D) Mode$
›Reveal solutionSolution
The median is the positional average determined by the central 50% of the ordered data — not by the extreme values.

Step 1 — What each measure uses

Geometric mean and Harmonic mean are mathematical averages computed from all the observations.

Mode is the value with the highest frequency — fixed by the most concentrated value.

Median is a positional average: arrange the data and take the middle value, so it is governed by where the central items fall, splitting the series into two 50% halves.

Step 2 — Match the description …
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CA Foundation 2025Set may-20251 markMCQ
$Q. Find out the mode from the following data : 100, 110, 125, 225, 325, 125, 90, 80, 455, 375, 125 (A) 325 (B) 110 (C) 455 (D) 125$
›Reveal solutionSolution

The most frequently occurring value is 125 (it appears 3 times), so the mode is 125.

Step 1 — Tally the frequencies

Data: 100, 110, 125, 225, 325, 125, 90, 80, 455, 375, 125.

Value Times it appears
125 3
every other value (100, 110, 225, 325, 90, 80, 455, 375) 1 each

Step 2 — Pick the mode

The mode is the value with the highest frequency. Here 125 occurs three times, more than any other value. …
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CA Foundation 2024Set sep-20241 markMCQ
$x$
›Reveal solutionSolution

Median class is 20–30 (where cumulative frequency crosses N/2 = 50); the formula gives 23.

Step 1 — Cumulative frequencies ( $N = 8+30+40+12+10 = 100$ )

Class $f$ Cumulative $f$
0–10 8 8
10–20 30 38
20–30 40 78
30–40 12 90
40–50 10 100

Step 2 — Locate the median class

$N/2 = 50$ . The cumulative frequency first exceeds 50 in the 20–30 class (cf jumps 38 → 78), so median class = 20–30.

Step 3 — Apply the median formula
$M = L + \frac{\tfrac{N}{2} - cf}{f}\times h$
with $L=20,\; cf=38,\; f=40,\; h=10$ : …
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CA Foundation 2023Set jun-20231 markMCQ
$Q. A Professor has given assignment to students in a Statistics class. A student Jagan computes the arithmetic mean and standard deviation for a set of 100 observations as 50 and 5 respectively. Later on, Sonali points out to Jagan that he has made a mistake in taking one observation as 100 instead of 50. What would be the correct mean if the wrong observation is corrected? (A) 50.5 (B) 49.9 (C) 49.5 (D) 50.1$
›Reveal solutionSolution

Corrected sum 4950 over 100 observations gives a mean of 49.5.

Step 1 — Recover the original total
$\sum x = n\bar{x} = 100 \times 50 = 5000$
Step 2 — Correct the wrong observation

Replace 100 by 50:
$\text{corrected sum} = 5000 - 100 + 50 = 4950$
Step 3 — New mean
$\bar{x}_{\text{corrected}} = \frac{4950}{100} = 49.5$
Watch out

The standard deviation of 5 is irrelevant to the mean correction — including it is a common trap. …

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CA Foundation 2023Set jun-20231 markMCQ
$Q. Find the mean of the following data Class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 91364623 (A) 23.7 (B) 35.7 (C) 39.7 (D) 43.7$
›Reveal solutionSolution

Σfx = 1535, Σf = 43, so mean = 1535/43 ≈ 35.7.

Step 1 — Class midpoints
$15,\;25,\;35,\;45,\;55,\;65,\;75$
Step 2 — Compute Σf and Σfx
$\sum f = 9+13+6+4+6+2+3 = 43$ $\sum fx = 135+325+210+180+330+130+225 = 1535$
Step 3 — Mean
$\bar{x} = \frac{\sum fx}{\sum f} = \frac{1535}{43} \approx 35.7$
Watch out

Divide by the total frequency (43), never by the number of class intervals (7) — that mistake inflates the mean badly. …

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CA Foundation 2023Set jun-20231 markMCQ
$Q. The Median of the following set of observations: 24, 18, 36, 42, 30, 28, 21, 29, 25, 33 is (A) 26.5 (B) 27.5 (C) 28.5 (D) 29.5$
›Reveal solutionSolution

Sorted, the middle pair (5th and 6th of 10) is 28 and 29 → median 28.5.

Step 1 — Sort the data
$18,\;21,\;24,\;25,\;28,\;29,\;30,\;33,\;36,\;42$
Step 2 — Locate the median position

With $n = 10$ (even), the median is the mean of the $\tfrac{n}{2}=5$ th and $\tfrac{n}{2}+1=6$ th values: 28 and 29.

Step 3 — Compute
$\text{Median} = \frac{28 + 29}{2} = 28.5$
Watch out

Never read the median off the unsorted list — always arrange the values in ascending order first. …

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CA Foundation 2023Set jun-20231 markMCQ
$f(x)$
›Reveal solutionSolution

Modal class 30-35; mode = 30 + (33/44)×5 = 33.75.

Step 1 — Identify the modal class

Highest frequency is 53, in class 30-35. So $L = 30$ , $f_1 = 53$ , previous $f_0 = 20$ , next $f_2 = 42$ , width $h = 5$ .

Step 2 — Apply the mode formula
$\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\times h$ $= 30 + \frac{53-20}{2(53)-20-42}\times 5 = 30 + \frac{33}{44}\times 5$
Step 3 — Compute
$= 30 + 0.75\times 5 = 30 + 3.75 = 33.75$
Watch out

$f_0$ and $f_2$ are the frequencies of the classes immediately before and after the modal class — mixing them up (e.g., using 42 as $f_0$ ) changes the answer. …

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CA Foundation 2023Set jun-20231 markMCQ
$Q. For a given data set: 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4; what is the median? (A) 8 (B) 6 (C) 4 (D) 9$
›Reveal solutionSolution

Sorted, the 8th of 15 values is the median = 6.

Step 1 — Sort the data

2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19.

Step 2 — Locate the middle position
$\text{Median position} = \frac{n+1}{2} = \frac{15+1}{2} = 8$
The 8th value is 6.

Watch out

4 occurs most often (the mode) and tempts as an answer — but the median is the positional middle, not the most frequent value. …

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CA Foundation 2022Set dec-20221 markMCQ
$Q. The median of the observations 42, 72, 35, 92, 67, 85, 72, 81, 51, 56 is (A) 69.5 (B) 72 (C) 64 (D) 61.5$
›Reveal solutionSolution

Sorted, the average of the 5th and 6th values is $(67+72)/2 = 69.5$ .

Step 1 — Arrange in ascending order
$35,\;42,\;51,\;56,\;67,\;72,\;72,\;81,\;85,\;92$
Step 2 — Locate the middle for even $n$

With $n=10$ , the median is the mean of the $\frac{n}{2}=5$ th and $\frac{n}{2}+1=6$ th observations.
$\text{5th}=67,\qquad \text{6th}=72$
Step 3 — Average them
$\text{Median} = \frac{67+72}{2} = \frac{139}{2} = 69.5$
Watch out

You must sort first. Reading the raw list's middle entries (67, 85) or picking 72 because it repeats gives a wrong median. …

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Value	Times it appears
125	3
every other value (100, 110, 225, 325, 90, 80, 455, 375)	1 each

Class	$f$	Cumulative $f$
0–10	8	8
10–20	30	38
20–30	40	78
30–40	12	90
40–50	10	100

←Previous questionQ14 — An aeroplane flies, along the four sides of a square at speeds of 100,…→Next questionQ16 — (a) A company has three machines A, B, C which produces 25%, 30% and 5…

$x$	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
$f(x)$	8	30	40	12	10
(A) 33
(B) 22.5
(C) 23
(D) 24

X	25-30	30-35	35-40	40-45	45-50	50-55
$f(x)$	20	53	42	42	41	43
(A) 31.75
(B) 30.75
(C) 33.75
(D) 35.75

Concept understanding — Measures of Central Tendency

How it works

Key properties

Quick example

Step 1 — recall the relation

Step 2 — substitute

Step 3 — solve

Step 1 — Count the fixed values

Step 2 — Use the mode condition

Step 1 — Criteria for an ideal average

Step 2 — Why the mean wins

Step 1 — What each measure uses

Step 2 — Match the description …

Step 1 — Tally the frequencies

Step 2 — Pick the mode

Step 1 — Cumulative frequencies ( $N = 8+30+40+12+10 = 100$ )

Step 2 — Locate the median class

Step 3 — Apply the median formula

Step 1 — Recover the original total

Step 2 — Correct the wrong observation

Step 3 — New mean

Step 1 — Class midpoints

Step 2 — Compute Σf and Σfx

Step 3 — Mean

Step 1 — Sort the data

Step 2 — Locate the median position

Step 3 — Compute

Step 1 — Identify the modal class

Step 2 — Apply the mode formula

Step 3 — Compute

Step 1 — Sort the data

Step 2 — Locate the middle position

Step 1 — Arrange in ascending order

Step 2 — Locate the middle for even $n$

Step 3 — Average them

Class interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	9	13	6	4	6	2	3
(A) 23.7
(B) 35.7
(C) 39.7
(D) 43.7

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